APPLY the Math

Transcription

APPLY the Math

258
NEL
Chapter
5
Trigonometry
and Acute
Triangles
GOALS
You will be able to
•
Use trigonometric ratios to determine
angles and sides in right triangles
•
Use technology to investigate and verify
the sine and cosine laws
•
Apply trigonometry to solve real-life
problems involving right and acute
triangles
? How might each person shown in
these photos use trigonometry in his
or her field?
NEL
259
5
Getting Started
WORDS You Need to Know
1. Match the term with the picture or example that best illustrates its
definition.
c) Pythagorean theorem
d) hypotenuse
a) acute angle
b) right angle
i)
iii)
e) opposite side
f ) adjacent side
v)
A
A
ii)
iv)
vi)
c
a
b
A
a2 1 b2 5c2
hypotenuse
u
opposite
SKILLS AND CONCEPTS You Need
Determining and Using the Primary Trigonometric
Ratios in Right Triangles
For a right angle triangle, the three primary trigonometric ratios are:
adjacent
sin u 5
opposite
hypotenuse
cos u 5
adjacent
hypotenuse
tan u 5
opposite
adjacent
The opposite side, adjacent side, and hypotenuse are determined by their
positions relative to a particular angle in the triangle.
EXAMPLE
State the primary trigonometric ratios for /C.
A
3 cm
B
260
5 cm
4 cm
C
NEL
Chapter 5
Getting Started
Solution
Communication
3
opposite
5
sin C 5
hypotenuse
5
4
adjacent
5
cos C 5
hypotenuse
5
A
hypotenuse
b = 5 cm
opposite
c = 3 cm
B
C
adjacent
a = 4 cm
tan C 5
B
12 cm
It is common practice to label
the vertices of a triangle with
upper case letters. The side
opposite each angle is labelled
with the corresponding lower
case letter.
opposite
3
5
hypotenuse
4
2. State the primary trigonometric ratios for /A.
a)
b)
A
8 cm
C
13 cm
5 cm
A
b
c
B
B
C
15 cm
Tip
a
C
17 cm
A
3. Calculate the measure of the indicated side or angle to the nearest unit.
a)
c)
A
Study
Aid
For help, see Essential Skills
Appendix, A-15.
4
13
A
8
10
b)
d)
b
10
NEL
50°
15°
25
d
Trigonometry and Acute Triangles
261
PRACTICE
Study
4. Use the Pythagorean theorem to determine the value of x to the
Aid
For help, see Essential Skills
Appendix.
nearest unit.
a)
A
x
B
4 cm
C
13 cm
Question
Appendix
4
A-4
5, 6, 7, 10, 11
b)
D
10 m
14 m
A-15
F
E
x
5. Using the triangles in question 4, determine the primary trigonometric
Tech
Support
For help setting a graphing
calculator to degree mode and
determining angles using the
inverse trigonometric keys, see
Technical Appendix, B-12.
ratios for each given angle. Then determine the angle measure to the
nearest degree.
a) /A
b) /D
c) /C
6. Use a calculator to evaluate to four decimal places.
a) sin 50°
b) cos 11°
c) tan 72°
7. Use a calculator to determine u to the nearest degree.
a) cos u 5 0.6820
b) tan u 5 0.1944
c) sin u 5 0.9848
8. Determine each unknown angle to the nearest degree.
a)
b)
38°
␪
␾
␪
12
5
␾
truck bed
3.1 m
1.4 m
␪
ground
9. Rudy is unloading a piano from his truck. The truck bed is 1.4 m
above ground, and he extended the ramp to a length of 3.1 m.
a) At what angle does the ramp meet the ground? Round your answer
to the nearest degree.
b) Rudy needs the angle to be 15° or less so that he can control the
piano safely. How much more should he extend the ramp? Round
your answer to the nearest tenth of a metre.
10. Explain, using examples, how you could use trigonometry to calculate
a) the measure of a side in a right triangle when you know one side
and one angle
b) the measure of an angle when you know two sides
262
Chapter 5
NEL
Getting Started
APPLYING What You Know
Landing at an Airport
For an airplane to land safely, the base of the clouds, or ceiling, above the
airport must be at least 600 m. Grimsby Airport has a spotlight that shines
perpendicular to the ground, onto the cloud above. At 1100 m from the
spotlight, Cory measures the angle between the spotlight’s vertical beam of
light, himself, and the illuminated spot on the base of the cloud to be 55°.
Cory’s eyes are 1.8 m above ground.
ceiling
55°
1.8 m
1100 m
?
With this cloud ceiling, is it safe for a plane to land at
Grimsby Airport?
A.
Sketch the right triangle that models this problem. Represent
the given information on your sketch and any other information
you know.
B.
Label the hypotenuse and the sides that are opposite and adjacent to
the 55° angle.
C.
Which trigonometric ratio (sine, cosine, or tangent) would you use to
solve the problem? Justify your choice.
D.
Use the trigonometric ratio that you chose in part C to write an
equation to calculate the height of the cloud ceiling. Solve the
equation and round your answer to the nearest metre.
E.
Add 1.8 m to account for Cory’s eyes being above ground.
Determine whether it is safe for the plane to land.
NEL
263
5.1
Applying the Primary
Trigonometric Ratios
GOAL
Use primary trigonometric ratios to solve real-life problems.
LEARN ABOUT the Math
angle of elevation
the angle between the
horizontal and the line of sight
when looking up at an object
Eric’s car alarm will sound if his car is disturbed, but it is designed to shut
off if the car is being towed at an angle of elevation of more than 15°.
Mike’s tow truck can lift a bumper no more than 0.88 m higher than the
bumper’s original height above ground. Eric’s car has these measurements:
• The front bumper is 3.6 m from the rear axle.
• The rear bumper is 2.8 m from the front axle.
angle of elevation
3.6 m
2.8 m
?
Will Mike be able to tow Eric’s car without the alarm sounding?
EXAMPLE
1
Selecting a strategy to solve a problem
involving a right triangle
Determine whether the car alarm will sound.
Jason’s Solution: Calculating the Angle of Elevation
3.6 m
sin u 5
264
Chapter 5
opposite
hypotenuse
0.88 m
If Eric’s car is towed from the front,
the car forms a right triangle with
a hypotenuse of 3.6 m. The side
opposite the angle of elevation, u,
is 0.88 m.
In a right triangle, the sine ratio
relates an angle to the opposite
side and the hypotenuse.
NEL
5.1
sin u 5
0.88
3.6
u 5 sin 21 a
0.88
b
3.6
u 8 14°
I used the inverse sine ratio to
calculate the angle.
I rounded to the nearest degree.
Since the angle is less than 15°, the car has
not been lifted enough to shut off the alarm.
2.8 m
0.88 m
sin u 5
opposite
hypotenuse
sin u 5
0.88
2.8
u 5 sin 21 a
0.88
b
2.8
u 8 18°
If the car is towed from the rear,
the car still forms a right triangle.
The opposite side is still 0.88, but
the hypotenuse is now 2.8 m.
To determine the angle of
elevation, u, I used the sine ratio,
since I knew the lengths of the
opposite side and the hypotenuse.
I used the inverse sine ratio to
calculate the angle.
I rounded to the nearest degree.
Since the angle is greater than 15°, the car
alarm will shut off.
For the car alarm to shut off, Mike should
tow Eric’s car from the back.
NEL
265
Monica’s Solution: Calculating the Minimum Height
3.6 m
15°
sin u 5
sin 15° 5
h
opposite
hypotenuse
For an angle of elevation of 15°,
I wanted to know the height the car
must be lifted to shut off the alarm.
I started with the front of the car
being lifted. I drew a right triangle
and labelled the height as h. I knew
an angle and the hypotenuse.
Side h is opposite the 15° angle, so
I used the sine ratio to calculate h.
h
3.6
1
3.6 3 sin 15° 5 3.6 3
h
3.6
I solved for h by multiplying both
sides of the equation by 3.6.
1
h 5 3.6 3 sin 15°
I used a calculator to evaluate and
rounded to the nearest hundredth.
h 8 0.93 m
Mike can’t raise the car high enough
from the front to shut off the alarm.
h
2.8 m
15°
sin 15° 5
266
Chapter 5
h
2.8
When the front of the car is lifted
15°, the front bumper is about
0.93 m above its original height.
This won’t work because Mike can
lift the bumper only 0.88 m.
I used the same method, but with
the rear of the car being lifted.
I used the sine ratio because I
knew the opposite side and the
hypotenuse.
NEL
5.1
1
2.8 3 sin 15° 5 2.8 3
h
2.8
1
h 5 2.8 3 sin 15°
I used a calculator to evaluate and
rounded to the nearest hundredth.
h 8 0.72 m
For the alarm to shut off, Mike should
tow Eric’s car from the rear.
When the rear of the car is lifted
15°, the rear bumper is about
0.72 m above its original height.
This will work, since Mike can lift
a car more than that.
Reflecting
A.
Compare the two solutions. How are they the same and how are
they different?
B.
Which solution do you prefer? Why?
C.
Could the cosine or tangent ratios be used instead of the sine ratio to
solve this problem? Explain.
APPLY the Math
EXAMPLE
2
Selecting the appropriate trigonometric
ratios to determine unknown sides
A hot-air balloon on the end of a taut 95 m rope rises from its platform. Sam, who
is in the basket, estimates that the angle of depression to the rope is about 55°.
a) How far, to the nearest metre, did the balloon drift horizontally?
b) How high, to the nearest metre, is the balloon above ground?
c) Viewed from the platform, what is the angle of elevation, to nearest degree,
to the balloon?
angle of depression
the angle between the
horizontal and the line of
sight when looking down
at an object
angle of depression
object
NEL
267
Kumar’s Solution
a)
I drew a sketch and labelled the
angle of depression as 55°.
I labelled the horizontal distance
the balloon drifted as d and the
height of the balloon as h. Relative
to the 55° angle, d is the adjacent
side and h is the opposite side.
d
55°
h
95 m
adjacent
cos u 5
hypotenuse
d
cos 55° 5
95
1
95 3 cos 55° 5 95 3
d
95
I used the cosine ratio to calculate
d because in a right triangle, cosine
relates an angle to the adjacent
I solved for d by multiplying both
sides of the equation by 95.
1
d 5 95 3 cos 55°
I used a calculator to evaluate.
d 8 54 m
The balloon drifted about 54 m
horizontally.
b)
sin 55° 5
h
95
1
95 3 sin 55° 5 95 3
h
95
I used the sine ratio to calculate h
because in a right triangle, sine
relates an angle to the opposite
1
h 5 95 3 sin 55°
268
Chapter 5
NEL
5.1
h 8 78 m
The balloon is about 78 m above
ground.
c)
I labelled the angle of elevation
to the balloon as u. The angle of
depression (55°) and u are
alternate angles between parallel
lines. So both angles are equal.
55°
The angle of elevation to the balloon
from the platform is 55°.
EXAMPLE
3
Solving a problem by using a trigonometric ratio
1
A wheelchair ramp is safe to use if it has a minimum slope of 12 and a
1
maximum slope of 5. What are the minimum and maximum angles of
elevation to the top of such a ramp? Round your answers to the nearest degree.
Nazir’s Solution
1
␪1
12
1
␪2
NEL
5
I drew a sketch of each ramp and labelled the angles of
elevation as u1 and u2. Slope is rise divided by run, so, in
both triangles, the rise is the vertical side and the run is the
horizontal side. Relative to u1 and u2, these sides are
opposite and adjacent.
269
tan u 5
tan u1 5
1
12
u1 5 tan21 a
opposite
adjacent
The tangent ratio relates an angle in a right triangle to
the opposite side and the adjacent side.
tan u2 5
1
b
12
1
5
1
u2 5 tan21 a b
5
I used the inverse tangent function on my
calculator to calculate each angle.
u1 8 5°
u2 8 11°
The angle of elevation to the top of the wheelchair ramp
must be between 5° and 11° to be safe.
In Summary
Key Idea
• The primary trigonometric ratios can be used to determine side lengths and
angles in right triangles. Which ratio you use depends on what you want to
calculate and what you know about the triangle.
Need to Know
• For any right ^ ABC, the primary trigonometric ratios for /A are:
sin A 5
opposite
hypotenuse
cos A 5
adjacent
hypotenuse
tan A 5
opposite
adjacent
C
hypotenuse
A
adjacent
opposite
B
• Opposite and adjacent sides are named relative to their positions to a particular
angle in a triangle.
• The hypotenuse is the longest side in a right triangle and is always opposite the
90° angle.
270
Chapter 5
NEL
5.1
CHECK Your Understanding
1. Use a calculator to evaluate to four decimal places.
a) sin 15°
b) cos 55°
2. Use a calculator to determine the angle to the nearest degree.
a)
2
tan 21 a b
5
b) sin 21 (0.7071)
3. State all the primary trigonometric ratios for /A and /D. Then
determine /A and /D to the nearest degree.
a)
b)
C
5
A
11
D
3
E
6.9
13
F
B
4
4. For each triangle, calculate x to the nearest centimetre.
a)
b)
C
x
50°
9 cm
53°
B
Q
A
10 cm
P
R
x
PRACTISING
5. Determine all unknown sides to the nearest unit and all unknown interior
K
angles to the nearest degree.
a)
c)
A
x
8
P
10
z
y
B
15
d)
j
R
9
J
l
i
q
y
Q
C
G
b)
x
K
72°
10
x
j
L
60°
H
NEL
8
I
271
6. Predict the order, from tallest to shortest, of these famous landmarks.
Then use the given information to determine the actual heights to the
nearest metre.
a) Eiffel Tower, Paris, France
68 m from the base, the angle of elevation to the top is 78°.
b) Empire State Building, New York, New York
267 m from the base, the angle of elevation to the top is 55°.
c) Leaning Tower of Pisa, Pisa, Italy
The distance from a point on the ground to the tallest tip of the
tower is 81 m with an angle of elevation of 44°.
d) Big Ben’s clock tower, London, England
81 m from the base, the angle of elevation is 50°.
Eiffel Tower
Leaning Tower
of Pisa
7. Manpreet is standing 8.1 m from a flagpole. His eyes are 1.7 m above
ground. The top of the flagpole has an angle of elevation of 35°. How
tall, to the nearest tenth of a metre, is the flagpole?
Empire State
Building
Big Ben’s clock
tower
8. The CN Tower is 553 m tall. From a position on one of the Toronto
A
Islands 1.13 km away from the base of the tower, determine the angle
of elevation, to the nearest degree, to the top of the tower. Assume that
your eyes are 1.5 m above ground.
553 m
␪
1.5 m
1.13 km
9. Devin wants to estimate the slope of the road near his apartment
building. He uses a level that is 1.2 m long and holds it horizontally
with one end touching the ground and the other end 39 cm above
ground.
a) What is the slope of the road, to the nearest tenth, at this point?
b) What angle, to the nearest degree, would represent the slant of
the road?
10. A 200 m cable attached to the top of an antenna makes an angle of
37° with the ground. How tall is the antenna to the nearest metre?
11. An underground parking lot is being constructed 8.00 m below
ground level.
a) If the exit ramp is to rise at an angle of 15°, how long will the
ramp be? Round your answer to the nearest hundredth of a metre.
b) What horizontal distance, to the nearest hundredth of a metre, is
needed for the ramp?
272
Chapter 5
NEL
5.1
12. The pitch of a roof is the rise divided by the run. If the pitch is
T
greater than 1.0 but less than 1.6, roofers use planks fastened to
the roof to stand on when shingling it. If the pitch is greater than
1.6, scaffolding is needed. For each roof angle, what equipment
(scaffolding or planks), if any, would roofers need?
a) u 5 34°
b) u 5 60°
c) u 5 51°
roof
angle
rise
13. To estimate the width of a river near their school, Charlotte and Mavis
run
have a device to measure angles and a pole of known length. Describe
how they might calculate the width of the river with this equipment.
14. Ainsley’s and Caleb’s apartment buildings are exactly the same height.
Ainsley measures the distance between the buildings as 51 m and observes
that the angle of depression from the roof of her building to the bottom of
Caleb’s is about 64°. How tall, to the nearest metre, is each building?
64°
15. To use an extension ladder safely, the base must be 1 m out from the
wall for every 2 m of vertical height.
a) What is the maximum angle of elevation, to the nearest degree, to
the top of the ladder?
b) If the ladder is extended to 4.72 m in length, how high can it safely
reach? Round your answer to the nearest hundredth of a metre.
c) How far out from the wall does a 5.9 m ladder need to be? Round
your answer to the nearest tenth of a metre.
51 m
16. Martin is installing an array of solar panels 3.8 m high in his backyard.
C
The array needs to be tilted 60° from the ground. Municipal bylaws
restrict residents from having any secondary structures taller than 3.0 m.
Will Martin be able to build his array? Include calculations in your
explanation.
Extending
17. When poured into a pile, gravel will naturally form a cylindrical cone
with a slope of approximately 34°. If a construction foreman has room
only for a pile that is 85 m in diameter, how tall, to the nearest metre,
will the pile be?
18. Nalini and Jodi are looking at the top of the same flagpole. They are
standing in a line on the same side of the flagpole, 50.0 m apart. The
angle of elevation to the top of the pole is 11° from Jodi’s position and
7° from Nalini’s position. The girls’ eyes are 1.7 m above ground. For
each question, round your answer to the nearest tenth of a metre.
a) How tall is the flagpole?
b) How far is each person from the base of the flagpole?
c) If Nalini and Jodi were standing in a line on opposite sides of the
pole, how tall would the flagpole be? How far would each person
be from its base?
NEL
273
5.2
Solving Problems by Using
Right-Triangle Models
GOAL
Solve real-life problems by using combinations of primary
trigonometric ratios.
LEARN about the Math
A surveyor stands at point S and uses a laser transit and marking pole to
sight two corners, A and B, on one side of a rectangular lot. The surveyor
marks a reference point P on the line AB so that the line from P to S is
perpendicular to AB. Corner A is 150 m away from S and 34° east of P.
Corner B is 347 m away from S and 69° west of P.
N
Tip
Communication
W
For the highest degree of
accuracy, save intermediate
answers by using the memory
keys of your calculator. Round
only after you do the very last
calculation.
347 m
1
69°
34°
A
150 m
S
?
EXAMPLE
S
P
B
E
How long, to the nearest metre, is the lot?
Selecting a strategy to determine a distance
Calculate the length of the building lot to the nearest metre.
Leila’s Solution: Using the Sine Ratio
x
B
347 m
P
y
P
34°
69°
S
A
The length of the lot is BP 1PA. I sketched
the two right triangles that these sides
formed. I labelled BP as x and PA as y.
150 m
S
In each triangle, I knew an angle and the
hypotenuse. I needed to calculate the sides
(x and y) opposite each given angle. So I
used the sine ratio.
opposite
sin u 5
hypotenuse
sin 69° 5
x
347
sin 34° 5
x
347 3 sin 69° 5 347 3
347
1
1
274
y
150
y
150 3 sin 34° 5 150 3
150
1
1
x 5 347 3 sin 69°
y 5 150 3 sin 34°
x 8 323.952 408
y 8 83.878 935 52
Chapter 5
To solve for x, I multiplied both sides of the
first equation by 347.
To solve for y, I multiplied both sides of the
second equation by 150.
NEL
5.2
AB 5 BP 1 PA
5x1y
I added x and y to determine the
length of the lot.
5 323.952 408 1 83.878 935 52
8 408 m
The lot is about 408 m long.
Tony’s Solution: Using the Cosine Ratio and the Pythagorean Theorem
P
B
69°
347 m
A
34°
150 m
The length of the lot is BP 1 PA. PS is a
shared side of ^ BPS and ^ APS. If I knew
PS, then I could use the Pythagorean
theorem to determine BP and PA.
S
In ^ APS:
cos u 5
cos 34° 5
adjacent
hypotenuse
AS is the hypotenuse and PS is adjacent to
the 34° angle. So, to determine PS, I used
the cosine ratio.
PS
150
1
150 3 cos 34° 5 150 3
PS
150
To solve for PS, I multiplied both sides of
the equation by 150 and used a calculator
to evaluate.
1
PS 5 150 3 cos 34°
PS 8 124.356 m
In ^ APS:
In ^ BPS:
AS 5 PS 1 PA
2
2
2
150 2 5 124.356 2 1 PA 2
BS 2 5 PS 2 1 BP 2
347 2 5 124.356 2 1 BP 2
PA 2 5 150 2 2 124.356 2
BP 2 5 347 2 2 124.356 2
PA 2 5 7036
BP 2 5 104 945
PA 5 !7036
BP 5 !104 945
PA 8 83.880 867 9 m
BP 8 323.952 157 m
AB 5 BP 1 PA
5 83.880 867 9 1 323.952 157
Next, I used the Pythagorean theorem to
determine PA and BP .
I added BP and PA to determine the length
of the lot.
8 408 m
The lot is about 408 m long.
NEL
275
Reflecting
A.
Explain why the length of the lot cannot be determined directly using
either the Pythagorean theorem or the primary trigonometric ratios.
B.
How are Leila’s and Tony’s solutions the same? How are they different?
C.
Why did the surveyor choose point P so that PS would be
perpendicular to AB?
APPLY the Math
EXAMPLE
2
Using trigonometric ratios to
calculate angles
Karen is a photographer taking pictures of the Burlington Skyway bridge. She is
in a helicopter hovering 720 m above the bridge, exactly 1 km horizontally from
the west end of the bridge. The Skyway spans a distance of 2650 m from east
to west.
a) From Karen’s position, what are the angles of depression, to the nearest
degree, of the east and west ends of the bridge?
b) If Karen’s camera has a wide-angle lens that can capture 150°, can she get
the whole bridge in one shot?
Ali’s Solution
a)
1000 m
720 m
1650 m
west
east
B
␣
A
276
Chapter 5
␤
720 m
1650 m
1000 m
␣
D
␤
C
I sketched the position of
the helicopter relative to
the bridge. The horizontal
distance from the helicopter
to the west end of the
bridge is 1000 m. So I
subtracted 1000 from 2650
to determine the horizontal
distance from the helicopter
to the east end.
I drew a dashed line parallel
to the base of the triangle
and labelled the angles of
depression as a and b. In
^ ABD, I knew that
/BAD 5 a because /BAD
and a are alternate angles
between parallel lines. By the
same reasoning, /BCD 5 b.
NEL
5.2
tan u 5
tan a 5
opposite
adjacent
720
1000
a 5 tan 21 a
tan b 5
720
b
1000
720
1650
b 5 tan 21 a
a 8 35.8°
I knew the opposite and
adjacent sides in each right
triangle. So, to determine a
and b, I used the tangent ratio.
720
b
1650
b 8 23.6°
The angles of depression are about 36° (west end)
and about 24° (east end).
b)
␣
␤
␪
I used the inverse tangent
function on my calculator
to evaluate each angle.
To determine whether Karen
can get the whole bridge in
one shot, I needed to know u,
the angle between the east
and west ends of the bridge
from Karen’s perspective.
I subtracted a and b
from 180°.
u 5 180° 2 a 2 b
5 180° 2 35.8° 2 23.6°
5 120.6°
Since the angle from Karen’s perspective is
less than 150°, she will be able to get the
whole bridge in one shot.
EXAMPLE
3
Using trigonometry to determine
the area of a triangle
Elise’s parents have a house with a triangular front lawn
as shown. They want to cover the lawn with sod rather
than plant grass seed. How much would it cost to put
sod if it costs $13.75 per square metre?
13 m
40°
11 m
Tina’s Solution
B
13 m
h
A
40°
D
11 m
NEL
C
I drew a dashed line for the
height of the triangle and labelled
it as h. The formula for the area
1
of a triangle is A 5 b 3 h.
2
I knew the base, but I needed to
calculate h before calculating
the area.
277
sin u 5
sin 40 ° 5
opposite
hypotenuse
h
13
1
13 3 sin 40° 5 13 3
h
13
1
h 5 13 3 sin 40°
In right ÂBD, I knew the
hypotenuse and the side adjacent to
the 40° angle. The height is the side
opposite the 40° angle, so I used the
sine ratio.
To solve for h, I multiplied both sides
of the equation by 13 and used a
calculator to evaluate.
h 8 8.356 m
A5
5
1
b3h
2
1
(11 3 8.35)
2
I substituted the base and height
into the formula for the area of a
triangle.
8 45.958 m2
cost 5 45.958 3 13.75
8 $631.92
Elise’s parents would have to spend
$631.92 to sod their lawn.
EXAMPLE
4
I multiplied the area by the cost per
square metre to determine the cost
to sod the lawn. I rounded to the
nearest cent.
Solving a problem by using trigonometric ratios
A communications tower is some distance from the base of a 70 m high
building. From the roof of the building, the angle of elevation to the top of the
tower is 11.2°. From the base of the building, the angle of elevation to the top of
the tower is 33.4°. Determine the height of the tower and how far it is from the
base of the building. Round your answers to the nearest metre.
11.2°
70 m
33.4°
278
Chapter 5
NEL
5.2
Pedro’s Solution
E
h
11.2°
A
I drew a sketch and labelled the distance between
the tower and the building as d. The tower is more
than 70 m tall, so I labelled the extra height above
70 m as h.
D
70 m
33.4°
B
C
d
In ^ BEC:
In ^ AED:
In ^ BEC, d is adjacent to the 33.4° angle and
the opposite side is 70 1 h. In ^ AED, d is
adjacent to the 11.2° angle and the opposite
side is h. Since I knew the adjacent and opposite
sides in each right triangle, I used the tangent
ratio.
opposite
tan u 5
adjacent
tan 33.4° 5
h 1 70
d
1
d 3 tan 33.4° 5 d 3
tan 11.2° 5
h 1 70
d
h
d
1
d 3 tan 11.2° 5 d 3
1
1
d 3 tan 33.4° 5 h 1 70
d5
h
d
d 3 tan 11.2° 5 h
To solve for d in the first equation, I multiplied both
sides by d and then divided both sides by tan 33.4°.
To solve for h in the second equation, I multiplied
both sides by d.
h 1 70
tan 33.4°
d5
d 3 tan 11.2° 1 70
tan 33.4°
tan 33.4° 3 d 5 d 3 tan 11.2° 1 70
I substituted the second equation into the first and
solved for d. I multiplied both sides of the equation
by tan 33.4°. Then I simplified.
d(tan 33.4° 2 tan 11.2°) 5 70
0.461 373 177 9d 8 70
d5
70
0.461 373 177 9
8 151.721 000 2 m
h 5 151.721 000 2 3 tan 11.2°
To determine h, I substituted the value of d into
the second equation.
8 30 m
70 1 30 5 100 m
To determine the height of the tower, I added
70 to h.
The tower is about 100 m tall and about 152 m from the
base of the building.
NEL
279
In Summary
Key Idea
• If a situation involves calculating a length or an angle, try to represent the
problem with a right-triangle model. If you can, solve the problem by
using the primary trigonometric ratios.
Need to Know
• To calculate the area of a triangle, you can use the sine ratio to determine the
height. For example, if you know a, b, and /C, then
sin C 5
h
a
h 5 a sin C
The area of a triangle is
B
1
A5 b3h
2
A5
a
h
1
b(a sin C)
2
A
C
b
1. Calculate the area of each triangle to the nearest tenth of a square
centimetre.
a)
b)
A
C
A
5.5 cm
11.8 cm
30°
B
6.0 cm
C
7.6 cm
73°
B
T
2. A mountain is 780.0 m high. From points A and C, the angles of
elevation to the top of the mountain are 67° and 54° as shown at the
left. Explain how to calculate the length of a tunnel from A to C.
780.0 m
67°
A
3. Karen and Anna are standing 23 m away from the base of a 23 m high
54°
B
280
Chapter 5
C
house. Karen’s eyes are 1.5 m above ground and Anna’s eyes are 1.8 m
above ground. Both girls observe the top of the house and measure its
angle of elevation. Which girl will measure the greater angle of
elevation? Justify your answer.
NEL
5.2
PRACTISING
4. The angle of elevation from the roof of a 10 m high building to the top
K
of another building is 41°. The two buildings are 18 m apart at the base.
a) Which trigonometric ratio would you use to solve for the height of
the taller building? Why?
b) How tall, to the nearest metre, is the taller building?
41°
10 m
5. If the angle of elevation to the top of the pyramid of Cheops in Giza,
Egypt, is 17.5°, measured 348 m from its base, can you calculate the
height of the pyramid accurately? Explain your reasoning.
6. Darrin wants to lean planks of wood that are 1.5 m, 1.8 m, and 2.1 m
long against the wall inside his garage. If the top of a plank forms an
angle of less than 10° with the wall, the plank might fall over. If
the bottom of a plank sticks out more than 30 cm from the wall,
Darrin won’t have room to park his car. Will Darrin be able to store
all three planks in the garage? Justify your answer with calculations.
18 m
10°
10°
10°
2.1 m
1.8 m
1.5 m
7. Jordan is standing on a bridge over the Welland Canal. His eyes are
5.1 m above the surface of the water. He sees a cargo ship heading
straight toward him. From his position, the bow appears at an angle of
depression of 5° and the stern appears at an angle of depression of 1°.
For each question, round your answer to the nearest tenth of a metre.
a) What is the straight-line distance from the bow to Jordan?
b) What is the straight-line distance from the stern to Jordan?
c) What is the length of the ship from bow to stern?
70°
8. A searchlight is mounted at the front of a helicopter flying 125 m
T
5°
above ground. The angle of depression of the light beam is 70°.
An observer on the ground notices that the beam of light measures 5°.
How wide, to the nearest metre, is d, the spot on the ground?
d
NEL
281
9. An 18 m long ladder is leaning against the wall of a building. The top of
the ladder reaches a window 11 m above ground. If the ladder is tilted in
the opposite direction, without moving its base, the top of the ladder
can reach a window in another building that is 7 m above ground. How
far apart, to the nearest metre, are the two buildings?
18 m
11 m
7m
10. Kyle and Anand are standing on level ground on opposite sides of a
d
tree. Kyle measures the angle of elevation to the treetop as 35°. Anand
measures an angle of elevation of 30°. Kyle and Anand are 65 m apart.
Kyle’s eyes and Anand’s eyes are 1.6 m above ground. How tall, to the
nearest tenth of a metre, is the tree?
11. A tree branch 3 m above ground runs parallel to Noel’s garage and his
2.7 m
neighbour’s. Noel wants to attach a rope swing on this branch. The
garages are 4.2 m apart and the rope is 2.7 m long.
a) What is the maximum angle, measured from the perpendicular,
through which the rope can swing? Round your answer to the
nearest degree.
b) What is the maximum height above ground, to the nearest tenth
of a metre, of the end of the rope?
␪
3m
h
4.2 m
12. A regular hexagon has a perimeter of 50 cm.
a) Calculate the area of the hexagon to the nearest square centimetre.
b) The hexagon is the base of a prism of height 100 cm. Calculate the
volume, to the nearest cubic centimetre, and the surface area, to the
nearest square centimetre, of the prism. Recall that volume 5 area
of base 3 height. The surface area is the sum of the areas of all
faces of the prism.
13. Lucien wants to photograph the separation of the solid rocket boosters
A
20 000 m
␪
148
24°
181
19°
x
from the space shuttle. He is standing 14 500 m from the launch pad,
and the solid rocket boosters separate at 20 000 m above ground.
a) Assume that the path of the shuttle launch is perfectly vertical up
until the boosters separate. At what angle, to the nearest degree,
should Lucien aim his camera?
b) How far away is the space shuttle from Lucien at the moment the
boosters separate? Round your answer to the nearest metre.
14. Sven wants to determine the unknown side of the triangle shown at the
C
left. Even though it is not a right triangle, describe how Sven could use
primary trigonometric ratios to determine x.
15. A pendulum that is 50 cm long is moved 40° from the vertical. What
2.7 m
is the change in height from its initial position? Round your answer to
the nearest centimetre.
Extending
60° 100°
0.6 m
282
Chapter 5
16. Calculate the area of the triangle shown at the left to the nearest tenth
of a square metre.
NEL
5.3
Investigating and Applying the
Sine Law in Acute Triangles
YOU WILL NEED
GOAL
• dynamic geometry software
Verify the sine law and use it to solve real-life problems.
INVESTIGATE the Math
The trigonometric ratios sine, cosine, and tangent are defined only for right
triangles. In an oblique triangle, these ratios no longer apply.
?
In an oblique triangle, what is the relationship between a side
and the sine of the angle opposite that side?
A.
Use dynamic geometry software to construct any acute triangle.
B.
Label the vertices A, B, and C. Then name the sides a, b, and c as shown.
C.
Measure all three interior angles and all three sides.
D.
Using one side length and the angle opposite that side, choose
Calculate ... from the Measure menu to evaluate the ratio
side length
.
sin(opposite angle)
E.
Repeat part D for the other sides and angles. What do you notice?
F.
Drag any vertex of your triangle. What happens to the sides, angles,
and ratios? Now drag the other two vertices and explain.
G.
Express your findings
• in words
• with a mathematical relationship
I.
J.
a triangle (acute or obtuse) that
does not contain a right angle
Tech
Support
For help using dynamic
geometry software, see
Technical Appendix, B-21
and B-22.
sine law
in any acute triangle, the ratios
of each side to the sine of its
opposite angle are equal
Reflecting
H.
oblique triangle
C
Are the reciprocals of the ratios you found equal? Explain how you
know.
The relationship that you verified is the sine law. Why is this name
appropriate?
Does the sine law apply to all types of triangles (obtuse, acute, and
right)? Explain how you know.
b
a
A
c
B
b
c
a
5
5
sin A
sin B
sin C
or
sin B
sin A
sin C
5
5
a
c
b
NEL
283
APPLY the Math
EXAMPLE
1
Using the sine law to calculate an
unknown length
Determine x to the nearest tenth of a centimetre.
55°
9.2 cm
70°
x
Barbara’s Solution
The triangle doesn’t have a 90°
angle. So it isn’t convenient to
use the primary trigonometric
ratios.
9.2
x
5
sin 55°
sin 70°
I chose the sine law because
I knew two angles and the side
opposite one of those angles.
I wrote the ratios with the sides
in the numerators to make the
calculations easier.
1
sin 55° 3
x
9.2
5 sin 55° 3
sin 55°
sin 70°
To solve for x, I multiplied both
sides of the equation by sin 55°.
1
x 5 sin 55° 3
9.2
sin 70°
x 8 8.0 cm
The unknown side of the triangle is about
8.0 cm long.
284
Chapter 5
NEL
5.3
EXAMPLE
2
Selecting the sine law as a strategy to
calculate unknown angles
Determine all the interior angles in ^PQR.
Round your answers to the nearest degree.
Q
112 cm
98 cm
124°
P
R
Tom’s Solution
/QPR 5 180° 2 124°
5 56°
sin 56°
sin R
5
112
98
/QPR and 124° add up to 180°.
So, to determine /QPR, I
subtracted 124° from 180°.
^ PQR doesn’t have a 90° angle.
So I chose the sine law, since I
knew two sides and an angle
opposite one of those sides.
I wrote the ratios with the angles
in the numerators to make it
easier to solve for /R.
1
112 3
sin R
sin 56°
5 112 3
112
98
1
sin R 5 112 3
To solve for /R, I multipled both
sin 56°
98
I used the inverse sine function
on a calculator to evaluate.
/R 8 71°
/Q 5 180° 2 /QPR 2 /R
5 180° 2 56° 2 71°
All three interior angles add
up to 180°. So I subtracted /R
and /QPR from 180° to
determine /Q.
5 53°
The interior angles in ^PQR are 56°, 53°,
and 71°.
NEL
285
EXAMPLE
3
Solving a problem by using the sine law
A wall that is 1.4 m long has started to lean and now makes an angle of 80°
with the ground. A 2.0 m board is jammed between the top of the wall and the
ground to prop the wall up. Assume that the ground is level.
a) What angle, to the nearest degree, does the
board make with the ground?
b) What angle, to the nearest degree, does the
2.0 m
board make with the wall?
1.4 m
c) How far, to the nearest tenth of a metre, is
the board from the base of the wall?
80°
␪
Isabelle’s Solution
This triangle doesn’t have a 90°
angle. So I chose the sine law,
since I knew two sides and an
angle opposite one of those
sides.
sin u
sin 80°
5
1.4
2.0
a)
1
1.4 3
sin u
sin 80°
5 1.4 3
1.4
2.0
1
sin u 5 1.4 3
To solve for u, I multiplied both
sin 80°
2.0
u 5 sin21 a1.4 3
sin 80°
b
2.0
I used the inverse sine ratio
on a calculator to determine
the angle.
u 8 44°
The board makes an angle of about
44° with the ground.
b) 180° 2 80° 2 44° 5 56°
The board makes an angle of about
56° with the wall.
The interior angles add up to
180°. So I subtracted the two
known angles from 180°.
c)
56°
2.0 m
1.4 m
80°
I labelled the distance along the
ground between the wall and
the board as d.
44°
d
286
Chapter 5
NEL
5.3
1
sin 56° 3
2.0
d
5
sin 56°
sin 80°
Then I used the sine law.
d
2.0
5 sin 56° 3
sin 56°
sin 80°
To solve for d, I multiplied
both sides of the equation by
sin 56° and used a calculator
to evaluate.
1
d 5 sin 56° 3
2.0
sin 80°
d 8 1.7 m
The board is about 1.7 m from the base of the wall.
In Summary
Key Idea
• The sine law states that, in any acute ^ ABC, the ratios of each side to the sine
of its opposite angle are equal.
a
b
c
5
5
sin A
sin B
sin C
C
or
b
sin B
sin C
sin A
5
5
a
b
c
a
A
c
B
Need to Know
• The sine law can be used only when you know
• two sides and the angle opposite a known side or
•
two angles and any side
or
NEL
287
1. a) Given ^XYZ at the left, label the sides with lower case letters.
b) State the sine law for ^XYZ.
Y
2. Solve each equation. Round x to the nearest tenth of a unit and u to
the nearest degree.
11.6
x
5
a)
sin 22°
sin 71°
29.2
13.1
5
b)
sin u
sin 65°
X
Z
3. Use the sine law to calculate b to the nearest centimetre and /D to the
nearest degree.
a)
D
b)
A
8 cm
8 cm
70°
B
35°
C
48°
E
Q
F
PRACTISING
4. Given ^RQS at the left, determine q to the nearest centimetre.
7 cm
K
5. Archimedes Avenue and Bernoulli Boulevard meet at an angle of 45°
80°
45°
R
q
S
near a bus stop. Riemann Road intersects Archimedes Avenue at an
angle of 60°. That intersection is 112 m from the bus stop.
a) At what angle do Riemann Road and Bernoulli Boulevard meet?
Round your answer to the nearest degree.
b) How far, to the nearest metre, is the intersection of Riemann Road
and Bernoulli Boulevard from the bus stop?
bus
stop
Bernoulli Blvd.
45°
112 m
Archimedes Ave.
288
Chapter 5
Riemann Rd.
60°
NEL
5.3
6. In ÂBC, two sides and an angle are given. Determine the value of
/C to the nearest degree and the length of b to the nearest tenth
of a centimetre.
a) a 5 2.4 cm, c 5 3.2 cm, /A 5 28°
b) a 5 9.9 cm, c 5 11.2 cm, /A 5 58°
c) a 5 8.6 cm, c 5 9.4 cm, /A 5 47°
d) a 5 5.5 cm, c 5 10.4 cm, /A 5 30°
7. An isosceles triangle has two 5.5 cm sides and two 32° angles.
T
a) Calculate the perimeter of the triangle to the nearest tenth
of a centimetre.
b) Calculate the area of the triangle to the nearest tenth of a
square centimetre.
8. Solve each triangle. Round each length to the nearest centimetre and
each angle to the nearest degree.
A
a)
Tip
To solve a triangle, determine
the measures of all unknown
sides and angles.
Q
c)
Communication
8 cm
70°
B
7 cm
35°
C
N
b)
80°
45°
R
S
D
d)
5 cm
L
28°
8 cm
M
48°
E
9 cm
F
9. Solve each triangle. Round each length to the nearest tenth of a unit
and each angle to the nearest degree.
a) ÂBC: a 5 10.3, c 5 14.4, /C 5 68°
b) ^DEF: /E 5 38°, /F 5 48°, f 5 15.8
c) ^GHJ: /G 5 61°, g 5 5.3, j 5 3.1
d) ^KMN: k 5 12.5, n 5 9.6, /N 5 42°
e) ^PQR: p 5 1.2, r 5 1.6, /R 5 52°
f ) ^XYZ: z 5 6.8, /X 5 42°, /Y 5 77°
NEL
289
10. Toby uses chains and a winch to lift engines at his father’s garage. Two
2.8 m
␪2
␪1
1.9 m
hooks in the ceiling are 2.8 m apart. Each hook has a chain hanging
from it. The chains are of length 1.9 m and 2.2 m. When the ends
of the chains are attached, they form an angle of 86°. In this
configuration, what acute angle, to the nearest degree, does each
chain make with the ceiling?
2.2 m
86°
11. Betsy installed cordless phones in the student centre at points A, B,
A
and C as shown. Explain how you can use the given information to
determine which two phones are farthest apart.
C
65°
289 m
51°
A
B
12. Tom says that he doesn’t need to use the sine law because he can
C
always determine a solution by using primary trigonometric ratios. Is
Tom correct? What would you say to Tom to convince him to use the
sine law to solve a problem?
Extending
13. Two angles in a triangle measure 54° and 38°. The longest side of the
triangle is 24 cm longer than the shortest side. Calculate the length, to
the nearest centimetre, of all three sides.
B
14. Use the sine law to show why the longest side of a triangle must be
opposite the largest angle.
A
15. A triangular garden is enclosed by a fence. A dog is on a 5 m leash
tethered to the fence at point P, 6.5 m from point C, as shown at the
left. If /ACB 5 41°, calculate the total length, to the nearest tenth of
a metre, of fence that the dog can reach.
P
41°
16. Tara built a sculpture in the shape of a huge equilateral triangle of side
6.5 m
C
290
Chapter 5
length 4.0 m. Unfortunately, the ground underneath the sculpture
was not stable, and one of the vertices of the triangle sank 55 cm into
the ground. Assume that Tara’s sculpture was built on level ground
originally.
a) What is the length, to the nearest tenth of a metre, of the two
exposed sides of Tara’s triangle now?
b) What percent of the triangle’s surface area remains above ground?
Round your answer to the nearest percent.
NEL
5
Mid-Chapter Review
FREQUENTLY ASKED Questions
Q:
What are the primary trigonometric ratios, and how do you
use them?
A:
Given ^ ABC, the primary trigonometric ratios for /A are:
sin A 5
opposite
hypotenuse
cos A 5
adjacent
hypotenuse
tan A 5
opposite
adjacent
C
hypotenuse
opposite
A
B
adjacent
Study
Aid
• See Lesson 5.1,
Examples 1, 2, and 3 and
Lesson 5.2, Examples 1 to 4.
• Try Mid-Chapter Review
Questions 1 to 8.
To calculate an angle or side using a trigonometric ratio:
• Label the sides of the triangle relative to either a given angle or the
one you want to determine.
• Write an equation that involves what you are trying to find by using
the appropriate ratio.
• Solve your equation.
Q:
What is the sine law, and when can I use it?
A:
In any acute triangle, the ratios of a side to the sine of its opposite
angle are equal.
C
a
b
c
5
5
sin A
sin B
sin C
b
a
Study
Aid
• See Lesson 5.3,
Examples 1, 2, and 3.
• Try Mid-Chapter Review
Questions 9, 10, and 11.
or
A
c
B
sin B
sin A
sin C
5
5
c
a
b
• The sine law can be used only when you know
• two sides and the angle opposite a known side or
•
two angles and any side
or
NEL
291
PRACTICE Questions
Lesson 5.1
b) The beam reflects off the pool surface and
strikes a wall 4.5 m away from the reflection
point. The angle the beam makes with the
pool is exactly the same on either side of the
reflection point. Calculate how far up the
wall, to the nearest tenth of a metre, the
spotlight will appear.
1. From a spot 25 m from the base of the
Peace Tower in Ottawa, the angle of elevation to
the top of the flagpole is 76°. How tall, to the
nearest metre, is the Peace Tower, including
the flagpole?
3. Solve each triangle. Round each length to the
nearest tenth of a unit and each angle to the
nearest degree.
a)
W
y
x
X
63°
9.6 cm
b)
Y
K
72.3 cm
j
2. A spotlight on a 3.0 m stand shines on the
J
surface of a swimming pool. The beam of light
hits the water at a point 7.0 m away from the
base of the stand.
68.8 cm
L
c)
n
L
M
57°
m
3.0 m
7.0 m
20.3 m
N
d) G
i
a) Calculate the angle the beam makes with the
pool surface. Round your answer to the
nearest degree.
H
24.5 m
12.8 m
I
4. The sides of an isosceles triangle are 12 cm,
12 cm, and 16 cm. Use primary trigonometric
ratios to determine the largest interior angle to
the nearest degree. (Hint: Divide the triangle
into two congruent right triangles.)
292
Chapter 5
NEL
Mid-Chapter Review
Lesson 5.2
What is the angle of depression, to the
nearest degree, from Martin’s eyes to the car?
b) What is the straight-line distance, to the
nearest metre, from Martin to his car?
a)
5. The mainsail of the small sailboat shown below
measures 2.4 m across the boom and 4.4 m up
the mainmast. The boom and mast meet at an
angle of 74°. What is the area of the mainsail to
the nearest tenth of a square metre?
mainmast
boom
6. A coast guard boat is tracking two ships using
radar. At noon, the ships are 5.0 km apart and
the angle between them is 90°. The closest ship
is 3.1 km from the coast guard boat. How far, to
the nearest tenth of a kilometre, is the other ship
from the coast guard boat?
Lesson 5.3
9. Solve each triangle. Round each length to the
nearest unit and each angle to the nearest
degree.
a) ^DEF: /D 5 67°, /F 5 42°, e 5 25
b) ^PQR: /R 5 80°, /Q 5 49°, r 5 8
c) ÂBC: /A 5 52°, /B 5 70°, a 5 20
d) ^XYZ: /Z 5 23°, /Y 5 54°, x 5 16
10. From the bottom of a canyon, Rita stands 47 m
directly below an overhead bridge. She estimates
that the angle of elevation of the bridge is about
35° at the north end and about 40° at the south
end. For each question, round your answer to
the nearest metre.
a) If the bridge were level, how long would it be?
b) If the bridge were inclined 4° from north to
south, how much longer would it be?
4°
south
north
x
3.1 km
47 m
35°
40°
5.0 km
11. The manufacturer of a reclining lawn chair is
7. An overhead streetlight can illuminate a circular
area of diameter 14 m. The light bulb is 6.8 m
directly above a bike path. Determine the angle
of elevation, to the nearest degree, from the edge
of the illuminated area to the light bulb.
6.8 m
planning to cut notches on the back of the chair
so that you can recline at an angle of 30° as
shown.
a) What is the measure of /B, to the nearest
degree, for the chair to be reclined at the
proper angle?
b) Determine the distance from A to B to the
nearest centimetre.
B
␪
8. Martin’s building is 105 m high. From the roof,
A
35 cm
30°
65 cm C
he spots his car in the parking lot. He estimates
that it is about 70 m from the base of the
building.
NEL
293
5.4
YOU WILL NEED
• dynamic geometry software
Investigating and Applying the
Cosine Law in Acute Triangles
GOAL
Verify the cosine law and use it to solve real-life problems.
INVESTIGATE the Math
The sine law is useful only when you know two sides and an angle opposite
one of those sides or when you know two angles and a side. But in the
triangles shown, you don’t know this information, so the sine law cannot
be applied.
P
B
3.2 km Q
7 cm
5.9 km
6.2 km
43°
A
?
Tech
Support
For help using dynamic
geometry software, see
Technical Appendix, B-21
and B-22.
294
Chapter 5
5 cm
C
R
How can the Pythagorean theorem be modified to relate the
sides and angles in these types of triangles?
A.
Use dynamic geometry software to construct any acute triangle.
B.
Label the vertices A, B, and C. Then name the sides a, b, and c as shown.
C.
Measure all three interior angles and all three sides.
D.
Drag vertex C until /C 5 90°.
E.
What is the Pythagorean relationship for this triangle? Using the
measures of a and b, choose Calculate ... from the Measure menu to
determine a 2 1 b 2. Use the measure of c and repeat to determine c 2.
NEL
5.4
F.
Does a 2 1 b 2 5 c 2? If not, are they close? Why might they be off by a
little bit?
G.
Move vertex C farther away from AB to create an acute triangle. How
does the value of a 2 1 b 2 compare with that of c 2?
H.
Using the measures of a 2 1 b 2 and c 2, choose Calculate ... from the
Measure menu to determine a 2 1 b 2 2 c 2. How far off is this value
from the Pythagorean theorem? Copy the table shown and record your
observations.
Triangle
a
b
c
/C
a2 1 b2
c2
a2 1 b2 2 c2
1
2
3
4
5
I.
Move vertex C to four other locations and record your observations.
Make sure that one of the triangles has /C 5 90°, while the rest
are acute.
J.
Use a calculator to determine 2 ab cos C for each of your triangles. Add
this column to your table and record these values. How do they
compare with the values of a 2 1 b 2 2 c 2?
K.
Based on your observations, modify the Pythagorean theorem to relate
c 2 to a 2 1 b 2.
Reflecting
L.
How is the Pythagorean theorem a special case of the relationship you
found? Explain.
M. Based on your observations, how did the value of /C affect the value
of a 2 1 b 2 2 c 2?
N.
Explain how you would use the cosine law to relate each pair of values
in any acute triangle.
• a 2 to b 2 1 c 2
• b 2 to a 2 1 c 2
cosine law
in any acute ^ ABC,
c2 5 a2 1 b2 2 2 ab cos C
C
b
a
A
c
B
NEL
295
APPLY the Math
EXAMPLE
1
Selecting the cosine law as a strategy to
calculate an unknown length
Determine the length of c to the nearest centimetre.
B
7 cm
c
43°
A
5 cm
C
Anita’s Solution
a 5 7 cm
b 5 5 cm
/C 5 43°
c 2 5 a 2 1 b 2 2 2 ab cos C
contained angle
the acute angle between two
known sides
c 2 5 72 1 52 2 2(7) (5)cos 43°
c 5 "72 1 52 2 2(7) (5)cos 43°
I can’t be certain that the triangle
has a 90° angle, and I can’t use the
sine law with the given information.
I chose the cosine law since I knew
two sides and the contained
angle, /C.
To solve for c, I took the square
root of both sides of the equation.
c 8 5 cm
Side c is about 5 cm long.
EXAMPLE
2
Selecting the cosine law as a strategy to calculate an unknown angle
Determine the measure of /R to the nearest degree.
P
3.2 km Q
5.9 km
6.2 km
R
296
Chapter 5
NEL
5.4
Kew’s Solution
p 5 5.9 km
I knew three sides and no angles.
q 5 6.2 km
r 5 3.2 km
r 2 5 p 2 1 q 2 2 2 pq cos R
3.22 5 5.92 1 6.22 2 2(5.9) (6.2)cos R
I chose the cosine law. I wrote the
formula in terms of /R.
3.22 2 5.92 2 6.22 5 22(5.9) (6.2)cos R
1
2 2(5.9) (6.2)
3.22 2 5.92 2 6.22
bcos R
5a
22(5.9) (6.2)
2 2(5.9) (6.2)
To solve for /R, I divided both sides of
the equation by 22(5.9) (6.2).
1
3.22 2 5.92 2 6.22
5 cos R
22(5.9) (6.2)
I used the inverse cosine function on
a calculator to evaluate.
/R 8 31°
The measure of /R is about 31°.
EXAMPLE
3
Solving a problem by using the cosine law
Ken’s cell phone detects two transmission antennas, one 7 km away and the
other 13 km away. From his position, the two antennas appear to be separated
by an angle of 80°. How far apart, to the nearest kilometre, are the two antennas?
Chantal’s Solution
distance between the two antennas
as d.
d
E
F
13 km
80°
7 km
D
NEL
297
e 5 7 km
The triangle doesn’t have a 90°
angle. I knew two sides and the
contained angle, so I chose the
cosine law.
f 5 13 km
/D 5 80°
d 2 5 e 2 1 f 2 2 2 ef cos D
d 2 5 72 1 132 2 2(7) (13)cos 80°
d 5 "72 1 132 2 2(7) (13)cos 80°
I wrote the formula in terms of /D.
To solve for d, I took the square
root of both sides of the equation.
d 8 14 km
The two antennas are about 14 km apart.
In Summary
Key Idea
C
• The cosine law states that in any
^ ABC,
b
a 5 b 1 c 2 2 bc cos A
2
2
2
b 5 a 1 c 2 2 ac cos B
2
2
2
a
A
c 2 5 a2 1 b 2 2 2 ab cos C
c
B
Need to Know
• The cosine law can be used only when you know
• two sides and the contained angle or
•
all three sides
• The contained angle in a triangle is the angle between two known sides.
298
Chapter 5
NEL
5.4
1. i) In which triangle is it necessary to use the cosine law to calculate
the third side? Justify your answer.
ii) State the formula you would use to determine the length of side b
in ÂBC.
a)
B
3 cm
b)
E
55°
A
8m
2 cm
b
f
C
F
25°
11 m
D
2. Determine the length of each unknown side to the nearest tenth
of a centimetre.
a)
b) A
A
9.8 cm
C
11.8 cm
12.6 cm
73°
40°
B
C
8.6 cm
B
3. Determine each indicated angle to the nearest degree.
a)
b)
D
A
?
5.5 cm
12 cm
10 cm
E
?
B
NEL
5.3 cm
8 cm
5.0 cm
F
C
299
PRACTISING
4. Determine each indicated length to the nearest tenth of a centimetre.
K
b) R
c)
X
L
a)
7.5 cm
?
25.6 cm
?
120°
85°
N
S
20°
?
Y
35°
21°
T
Z
18.7 cm
11.2 cm
M
5. Solve each triangle. Round each length to the nearest tenth of a
centimetre and each angle to the nearest degree.
a)
^ ABC: /A 5 68°, b 5 10.1 cm, c 5 11.1 cm
b) ^ DEF: /D 5 52°, e 5 7.2 cm, f 5 9.6 cm
c)
^ HIF: /H 5 35°, i 5 9.3 cm, f 5 12.5 cm
d) ^PQR: p 5 7.5 cm, q 5 8.1 cm, r 5 12.2 cm
6. A triangle has sides that measure 5 cm, 6 cm, and 10 cm. Do any of
the angles in this triangle equal 30°? Explain.
7. Two boats leave Whitby harbour at the same time. One boat heads
A
19 km to its destination in Lake Ontario. The second boat heads
on a course 70° from the first boat and travels 11 km to its
destination. How far apart, to the nearest kilometre, are the boats
when they reach their destinations?
Whitby
70°
11 km
19 km
8. Louis says that he has no information to use the sine law to solve
G
h
f
39°
F
56°
71.7 m
300
^FGH shown at the left and that he must use the cosine law instead.
Is he correct? Describe how you would solve for each unknown side
and angle.
Chapter 5
H
NEL
5.4
9. Driving a snowmobile across a frozen lake, Sheldon starts from the
most westerly point and travels 8.0 km before he turns right at an
angle of 59° and travels 6.1 km, stopping at the most easterly point of
the lake. How wide, to the nearest tenth of a kilometre, is the lake?
10. What strategy would you use to calculate angle u in ^PQR shown at the
P
18 cm
11. A clock with a radius of 15 cm has an 11 cm minute hand and a 7 cm
T
Q
␪
right? Justify your choice and then use your strategy to solve for u to the
nearest degree.
75°
hour hand. How far apart, to the nearest centimetre, are the tips of the
hands at each time?
a) 3:30 p.m.
b) 6:38 a.m.
9 cm
R
12. Use the triangle shown at the right to create a problem that involves its
C
side lengths and interior angles. Then describe how to determine d.
25 m
78°
35 m
Extending
13. Two observers standing at points X and Y are 1.7 km apart. Each
person measures angles of elevation to two balloons, A and B, flying
overhead as shown. For each question, round your answer to the
nearest tenth of a kilometre.
d
A
B
80°
20°
X
25°
1.7 km
75°
Y
a) How far is balloon A from point X ? From point Y ?
b) How far is balloon B from point X ? From point Y ?
c) How far apart are balloons A and B?
NEL
301
5.5
Solving Problems by Using
Acute-Triangle Models
GOAL
Solve problems involving the primary trigonometric ratios and the
sine and cosine laws.
N
LEARN ABOUT the Math
Toronto
Steve leaves the marina at Jordan on a 40 km sailboat race across
Lake Ontario intending to travel on a bearing of 355°, but an early
morning fog settles. By the time it clears, Steve has travelled 32 km on
a bearing of 22°.
40 km
22°
32 km
?
In which direction must Steve head to reach the finish line?
EXAMPLE
355°
1?
Solving a problem by using the sine
and cosine laws
Jordan
Determine the direction in which Steve should head to reach the finish line.
bearing
Liz’s Solution
the direction in which you have
to move in order to reach an
object. A bearing is a clockwise
angle from magnetic north. For
example, the bearing of the
lighthouse shown is 335°.
T
j
S
N
40 km
27°
E
335°
32 km
J
25°
/J 5 5° 1 22°
5 27°
s 5 40 km
t 5 32 km
j 2 5 s 2 1 t 2 2 2 st cos J
j 2 5 40 2 1 32 2 2 2(40) (32)cos 27°
302
Chapter 5
interior angle I needed to
calculate as u.
A bearing of 355° means that
Steve’s destination should have
been 5° west of north. But he
drifted 22° east of north. So I
added 5° and 22° to get /J.
I didn’t have information to be
able to use the sine law to
determine angle u. So I used the
cosine law to first calculate side j.
NEL
5.5
j 5 "40 2 1 32 2 2 2(40) (32)cos 27°
To solve for j, I took the square
root of both sides of the
equation.
j 8 18.521 km
sin 27°
sin u
5
40
18.521
1
40 3
I now had information to be
able to use the sine law.
sin u
sin 27°
5 40 3
40
18.521
1
sin u 5 40 3
sin 27°
18.521
u 5 sin21 a40 3
sin 27°
b
18.521
To solve for angle u, I multiplied
both sides of the equation
by 40.
I used the inverse sine function
on a calculator to evaluate.
u 8 79°
N
22°
T
j
79°
S
?
I now knew angle u, but I needed
to determine the bearing to T
from S. I sketched the triangle
with side t extended. I labelled
the angles I knew and the angle
I needed to determine.
22° t
J
bearing 5 22° 1 180° 1 79°
5 281°
Steve must head at a bearing of about
281° to reach the finish line.
The bearing is measured from
due north. So I added all the
angles from due north to side j.
I knew that the angle along the
extended side t is 180°.
Reflecting
A.
Why didn’t Liz first use the sine law in her solution?
B.
How is a diagram of the situation helpful? Explain.
C.
Is it possible to solve this problem without using the sine law or the
cosine law? Justify your answer.
NEL
303
APPLY the Math
2
EXAMPLE
Solving a problem by using primary trigonometric ratios
A ladder leaning against a wall makes an angle of 31° with the wall. The ladder
just touches a box that is flush against the wall and the ground. The box has a
height of 64 cm and a width of 27 cm. How long, to the nearest centimetre, is
the ladder?
31°
64 cm
27 cm
Denis’s Solution
A
x1
31°
B
The box and ladder form two right triangles.
I labelled the two parts of the ladder as x1 and x2.
The total length of the ladder is the sum of x1 and x2.
C
31°
64 cm
Since CE is parallel to AD, /ECF is equal to /BAC,
which is 31°.
x2
27 cm
D
F
E
In ^ ABC:
In ÂBC, BC 5 27 cm; in ^CEF, CE 5 64 cm.
In ^CEF:
opposite
sin u 5
hypotenuse
adjacent
cos u 5
hypotenuse
sin 31° 5
BC
x1
cos 31° 5
CE
x2
sin 31° 5
27
x1
cos 31° 5
64
x2
1
x1 3 sin 31° 5 x1 3
27
x1
1
x2 3 cos 31° 5 x2 3
304
Chapter 5
To solve for x1, I multiplied both sides of the
equation by x1 and divided both sides by sin 31°.
64
x2
1
1
x1 3 sin 31° 5 27
In each triangle, I knew a side and an angle and
I needed to calculate the hypotenuse. So I used
primary trigonometric ratios:
• sine, since BC is opposite the 31° angle
• cosine, since CE is adjacent to the 31° angle
To solve for x2, I multiplied both sides of the
equation by x2 and divided both sides by cos 31°.
x2 3 cos 31° 5 64
NEL
5.5
1
x1 3 sin 31 °
27
5
sin 31 °
sin 31°
1
1
64
x2 3 cos 31 °
5
cos 31 °
cos 31°
1
27
sin 31°
x1 8 52.423 308 71
64
cos 31°
x2 8 74.664 537 42
x1 5
x2 5
length 5 x1 1 x2
I added x1 and x2 to determine the length
of the ladder.
5 52.423 308 71 1 74.664 537 42
8 127 cm
The ladder is about 127 cm long.
EXAMPLE
3?
Selecting a strategy to calculate the area
of a triangle
Jim has a triangular backyard with side lengths of 27 m, 21 m, and 18 m.
His bag of fertilizer covers 400 m2. Does he have enough fertilizer?
Barbara’s Solution
B
c = 21 m
h
a = 18 m
A
C
I drew a sketch with the
longest side at the
bottom. I labelled all the
given information.
b = 27 m
a 2 5 b 2 1 c 2 2 2 bc cos A
18 5 27 1 21 2 2(27) (21)cos A
2
2
2
182 2 27 2 2 212 5 22(27) (21)cos A
1
2 2(27) (21)
182 2 27 2 2 212
5a
bcos A
22(27) (21)
2 2(27) (21)
1
cos A 5
18 2 272 2 212
22(27) (21)
2
/A 5 cos21 a
182 2 272 2 212
b
22(27) (21)
To determine the area,
I needed the height, h,
of the triangle. To
determine h, I had to
calculate an angle first,
so I chose /A. I used
the cosine law to
determine /A.
To solve for /A, I
divided both sides of
the equation by
22(27) (21).
I used the inverse cosine
function on a calculator
to evaluate.
/A 8 41.75°
NEL
305
sin A 5
opposite
hypotenuse
sin A 5
h
21
sin 41.75° 5
h
21
1
21 3 sin 41.75° 5 21 3
Since I knew /A,
I used a primary
trigonometric ratio to
calculate h.
To solve for h, I
multiplied both sides of
the equation by 21.
h
21
1
21 3 sin 41.75° 5 h
I used a calculator to
evaluate.
13.98 m 8 h
A5
1
b3h
2
I substituted the values
for b and h into the
formula for the area of a
triangle.
1
5 a b27 3 13.98
2
8 189 m2
Since 189 is less than half of 400, Jim has enough
fertilizer to cover his lawn twice and still have some
fertilizer left over.
EXAMPLE
4
Solving a problem to determine a perimeter
A regular octagon is inscribed in a circle of radius
15.8 cm. What is the perimeter, to the nearest
tenth of a centimetre, of the octagon?
B
C
A
15.8
cm
Shelley’s Solution
B
cm
15.8
A
306
Chapter 5
C
An octagon is made up of eight
identical triangles, each of which
is isosceles because two sides
are the same length (radii of
the circle).
NEL
5.5
/A 5 360° 4 8
A circle measures 360°. /A is 1
8 of
360°. So I divided 360° by 8 to
get /A.
5 45°
A
45°
15.8 cm
B
I labelled the base of ^BAC as a.
C
a
/A 1 /B 1 /C 5 180°
Since ^BAC is isosceles,
/B 5 /C. The sum of the interior
angles is 180°, so I calculated /B.
45° 1 2/B 5 180°
2/B 5 180° 2 45°
2/B 5 135°
/B 5 67.5°
b
a
5
sin A
sin B
a
15.8
5
sin 45°
sin 67.5°
1
a
15.8
5 sin 45° 3
sin 45 ° 3
sin 45 °
sin 67.5°
I knew an angle and the side
opposite that angle. So I used the
sine law to calculate a.
To solve for a, I multiplied both
sides of the equation by sin 45°.
1
a 5 sin 45° 3
15.8
sin 67.5°
a 8 12.09 cm
perimeter 5 12.09 3 8
8 96.7 cm
The perimeter of the octagon is about
96.7 cm.
NEL
The octagon is made up of eight
sides of length equal to a. So I
multiplied a by 8 to get the
perimeter.
307
In Summary
Key Idea
• The primary trigonometric ratios, the sine law, and the cosine law can be used to
solve problems involving triangles. The method you use depends on the information
you know about the triangle and what you want to determine.
Need to Know
• If the triangle in the problem is a right triangle, use the primary trigonometric ratios.
• If the triangle is oblique, use the sine law and/or the cosine law.
Given
Required to Find
Use
angle
sine law
side
sine law
side
cosine law
angle
cosine law
SSA
ASA
or
AAS
SAS
SSS
308
Chapter 5
NEL
5.5
1. For each triangle, describe how you would solve for the unknown side
or angle.
a)
b)
A
R
3.1
S
42°
R
c)
3.6
4.1
43.0
x
␪
2.2
␪
39°
C
5.2
Q
T
B
P
2. Complete a solution for each part of question 1. Round x to the
nearest tenth of a unit and u to the nearest degree.
B
PRACTISING
3. Determine the area of
K
30°
^ ABC, shown at the right, to the nearest
square centimetre.
50 cm
4. The legs of a collapsible stepladder are each 2.0 m long. What is the
maximum distance between the front and rear legs if the maximum angle
at the top is 40°? Round your answer to the nearest tenth of a metre.
5. To get around an obstacle, a local electrical utility must lay two
A
A
50°
sections of underground cable that are 371.0 m and 440.0 m long.
The two sections meet at an angle of 85°. How much extra cable is
necessary due to the obstacle? Round your answer to the nearest tenth
of a metre.
C
6. A surveyor is surveying three locations (M, N, and P) for new rides in
an amusement park around an artificial lake. /MNP is measured as
57°. MN is 728.0 m and MP is 638.0 m. What is the angle at M to
the nearest degree?
7. Mike’s hot-air balloon is 875.0 m directly above a highway. When he is
looking west, the angle of depression to Exit 81 is 11°. The exit
numbers on this highway represent the number of kilometres left
before the highway ends. What is the angle of depression, to the
nearest degree, to Exit 74 in the east?
11°
west
Exit 81
NEL
875.0 m
␪
east
Exit 74
309
8. A satellite is orbiting Earth 980 km above Earth’s surface. A receiving
dish is located on Earth such that the line from the satellite to the dish
and the line from the satellite to Earth’s centre form an angle of 24° as
shown at the left. If a signal from the satellite travels at 3 3 108 m>s,
how long does it take to reach the dish? Round your answer to the
nearest thousandth of a second.
24°
980 km
9. Three circles with radii of 3 cm, 4 cm, and 5 cm are touching each
other as shown. A triangle is drawn connecting the three centres.
Calculate all the interior angles of the triangle. Round your answers to
the nearest degree.
r = 4 cm
r = 3 cm
r = 5 cm
10. Given the regular pentagon shown at the left, determine its perimeter
T
C
to the nearest tenth of a centimetre and its area to the nearest tenth of
a square centimetre.
11. A 3 m high fence is on the side of a hill and tends to lean over. The
h = 4.5 cm
hill is inclined at an angle of 20° to the horizontal. A 6.3 m brace is to
be installed to prop up the fence. It will be attached to the fence at a
height of 2.5 m and will be staked downhill from the base of the fence.
What angle, to the nearest degree, does the brace make with the hill?
6.3 m
3m
20°
12. A surveyor wants to calculate the distance BC across a river. He selects
B
39°
52°
A
86.0 m
C
310
Chapter 5
a position, A, so that CA is 86.0 m. He measures /ABC and /BAC as
39° and 52°, respectively, as shown at the left. Calculate the distance
BC to the nearest tenth of a metre.
NEL
5.5
13. For best viewing, a document holder for people who work at
computers should be inclined between 61° and 65° (/ABC) .
A 12 cm support leg is attached to the holder 9 cm from the bottom.
Calculate the minimum and maximum angle u that the leg must make
with the holder. Round your answers to the nearest degree.
14. Match each method with a problem that can be solved by that
C
A
9 cm
12 cm
leg
B
method. Describe how each method could be used to complete
a solution.
Method
holder
C
Problems
Cosine law
Chris lives in a U-shaped building. From his window,
he sights Bethany’s window at a bearing of 328° and
Josef’s window at a bearing of 19°. Josef’s window is
54 m from Bethany’s and both windows are directly
opposite each other. How far is each window from
Chris’s window?
Sine law
When the Sun is at an angle of elevation of 41°,
Martina’s treehouse casts a shadow that is 11.4 m long.
Assuming that the ground is level, how tall is Martina’s
treehouse?
Primary
trigonometric
ratios
Ken walks 3.8 km west and then turns clockwise
65° before walking another 1.7 km. How far
does Ken have to walk to get back to where he started?
Extending
15. Two paper strips, each 2.5 cm wide, are laying across each other at an
angle of 27°, as shown at the right. What is the area of the overlapping
paper? Round your answer to the nearest tenth of a square centimetre.
27°
16. The diagram shows a roofing truss with AB parallel to CD. Calculate
the total length of wood needed to construct the truss. Round your
answer to the nearest metre.
B
12 m
D
A
40°
5m
C
E
17. Lucas takes a 10.0 m rope and creates a triangle with interior angles
of 30°, 70°, and 80°. How long, to the nearest tenth of a metre, is
each side?
NEL
311
Curious Math
Cycling Geometry
Competitive cyclists pay great attention to the geometry of their legs and of
their bikes. To provide the greatest leverage and achieve optimum output, a
cyclist’s knees must form a right angle with the pedals at the top of a stroke
and an angle of 165° at the bottom of a stroke.
165°
To achieve these conditions, cyclists adjust the seat height to give the
appropriate distances.
1. When Mark is standing, the distance from his hips to his knees is
44 cm and from his hips to the floor is 93 cm. When Mark’s knees
form each angle listed, how far are his hips from his heels? Round
your answers to the nearest centimetre.
a) 90°
b) 165°
2. On Mark’s bike, one pedal is 85 cm from the base of the seat. For
Mark to achieve optimum output, by how much should he raise
his seat? Round your answer to the nearest centimetre.
85 cm
3. The table below lists the measurements of other cyclists in Mark’s
riding club. Would these cyclists be able to ride Mark’s bike?
What seat height(s) would they require? Round your answers to the
nearest centimetre.
Name
312
Chapter 5
Hips to Knees
Hips to Floor
Terry
38 cm
81 cm
Colleen
45 cm
96 cm
Sergio
41 cm
89 cm
NEL
5
Chapter Review
FREQUENTLY ASKED Questions
Q:
What is the cosine law, and how do you use it to determine
angles and sides in triangles?
A:
The cosine law is a relationship that is true for all triangles:
C
b
a 2 5 b 2 1 c 2 2 2 bc cos A
a
A
c
Study
Aid
• See Lesson 5.4,
Examples 1, 2, and 3.
• Try Chapter Review
Questions 7 and 8.
b 2 5 a 2 1 c 2 2 2 ac cos B
B
c 2 5 a 2 1 b 2 2 2 ab cos C
If you don’t know a side and an angle opposite it, use the cosine law.
The sine law can be used only if you know a side and the angle
opposite that side. You can use the cosine law in combination with
the sine law and other trigonometric ratios to solve a triangle.
Q:
How do you know which strategy to use to solve
a trigonometry problem?
A:
Given
Use
A right triangle with any two pieces
of information
Primary trigonometric ratios:
B
A
C
A triangle with information
about sides and opposite angles
SSA
or
ASA
sin A 5
opposite
hypotenuse
cos A 5
adjacent
hypotenuse
tan A 5
opposite
adjacent
Study
Aid
• See Lesson 5.5,
Examples 1 to 4.
• Try Chapter Review
Questions 9 and 10.
The sine law:
b
c
a
5
5
sin A
sin B
sin C
or
sin B
sin C
sin A
5
5
a
b
c
AAS
or
A triangle with no information that
allows you to use the sine law
SSS
SAS
or
NEL
The cosine law:
a 2 5 b 2 1 c 2 2 2 bc cos A
b 2 5 a 2 1 c 2 2 2 ac cos B
c 2 5 a 2 1 b 2 2 2 ab cos C
313
PRACTICE Questions
Lesson 5.1
Lesson 5.3
1. Determine x to the nearest unit and angle u
to the nearest degree.
a)
each length to the nearest centimetre and each
angle to the nearest degree.
Q
S
x 75°
5. Use the sine law to solve each triangle. Round
a)
16 cm
A
8 cm
R
75°
T
b)
25°
B
b)
␪
N
18 cm
10 cm
6m
L
30°
M
c)
U
5m
C
Q
V
2. From Tony’s seat in the classroom, his eyes are
1.0 m above ground. On the wall 4.2 m away,
he can see the top of a blackboard that is 2.1 m
above ground. What is the angle of elevation, to
the nearest degree, to the top of the blackboard
from Tony’s eyes?
7 cm
R
85°
50°
S
d)
D
Lesson 5.2
3. A triangular garden has two equal sides 3.6 m
long and a contained angle of 80°.
a) How much edging, to the nearest metre,
is needed for this garden?
b) How much area does the garden cover?
Round your answer to the nearest tenth of
a square metre.
4. A Bascule bridge is usually built over water and has
two parts that are hinged. If each part is 64 m long
and can fold up to an angle of 70° in the upright
position, how far apart, to the nearest metre, are
the two ends of the bridge when it is fully open?
64 m
11 cm
E
52°
F
9 cm
6. A temporary support cable for a radio antenna is
110 m long and has an angle of elevation of 30°.
Two other support cables are already attached,
each at an angle of elevation of 70°. How long,
to the nearest metre, is each of the shorter cables?
64 m
70° 70°
110 m
30°
70° 70°
314
Chapter 5
NEL
Chapter Review
Lesson 5.4
8. A security camera needs to be placed so that
7. Use the cosine law to calculate each unknown
side length to the nearest unit and each
unknown angle to the nearest degree.
a)
A
5 cm
both the far corner of a parking lot and an entry
door are visible at the same time. The entry
door is 23 m from the camera, while the far
corner of the parking lot is 19 m from the
camera. The far corner of the parking lot is
17 m from the entry door. What angle of view
for the camera, to the nearest degree, is
required?
48°
B
C
6 cm
b)
T
23 m
3m
4m
U
17 m
5m
Lesson 5.5
V
c)
N
6 cm
60°
10 cm
M
Y
40°
9m
X
answers to the nearest degree and to the nearest
tenth of a centimetre.
a) ÂBC: /B 5 90°, /C 5 33°,
b 5 4.9 cm
b) ^DEF: /E 5 49°, /F 5 64°,
e 5 3.0 cm
c) ^GHI: /H 5 43°, g 5 7.0 cm,
i 5 6.0 cm
d) ^JKL: j 5 17.0 cm, k 5 18.0 cm,
l 5 21.0 cm
10. Two sides of a parallelogram measure 7.0 cm
9m
Z
NEL
9. Sketch and solve each triangle. Round your
L
d)
19 m
and 9.0 cm. The longer diagonal is 12.0 cm
long.
a) Calculate all the interior angles, to the
nearest degree, of the parallelogram.
b) How long is the other diagonal? Round
your answer to the nearest tenth of a
centimetre.
315
5
Chapter Self-Test
1. A 3 m ladder can be used safely only at an angle of 75° with the
horizontal. How high, to the nearest metre, can the ladder reach?
2. A road with an angle of elevation greater than 4.5° is steep for large
111 m
vehicles. If a road rises 61 m over a horizontal distance of 540 m, is the
road steep? Explain.
x
3. A surveyor has mapped out a property as shown at the left. Determine
y
the length of sides x and y to the nearest metre.
4. Solve each triangle. Round each length to the nearest centimetre and
78°
each angle to the nearest degree.
a)
134 m
B
18 cm
A
b)
E
78°
27°
28 cm
D
55°
F
15 cm
C
5. A 5.0 m tree is leaning 5° from the vertical. To prevent it from leaning
any farther, a stake needs to be fastened 2 m from the top of the tree at
an angle of 60° with the ground. How far from the base of the tree,
to the nearest metre, must the stake be?
6. A tree is growing vertically on a hillside that is inclined at an angle of
57°
7m
15°
15° to the horizontal. The tree casts a shadow uphill that extends 7 m
from the base of its trunk when the angle of elevation of the Sun is 57°.
How tall is the tree to the nearest metre?
7. Charmaine has planned a nature walk in the forest to visit four
B
stations: A, B, C, and D. Use the sketch shown at the left to calculate
the total length, to the nearest metre, of the nature trail, from A to B,
B to C, C to D, and D back to A.
271 m
41°
A
316
8. A weather balloon at a height of 117 m has an angle of elevation of
180 m
17°
247 m
D
Chapter 5
C
41° from one station and 62° from another. If the balloon is directly
above the line joining the stations, how far apart, to the nearest metre,
are the two stations?
NEL
5
Chapter Task
Crime Scene Investigator
mm
1
droplet A
2
3
4
droplet B
5
6
A great deal of mathematics and science is used in crime scene investigation.
For example, a blood droplet is spherical when falling. If the droplet strikes a
surface at 90°, it will form a circular spatter. If it strikes a surface at an angle,
the spatter spreads out.
droplet C
7
8
Suppose blood droplets, each 0.8 cm wide, were found on the ruler shown
at the right.
9
10
11
droplet D
12
A
B
AB = width of droplet
BC = spatter length
= angle of impact
dir
ec
t
mo ion o
tio
f
n
A
B
B.
C
Which droplet appears most circular? What is the length of that
droplet on the ruler, to the nearest tenth of a centimetre? What would
be the angle of impact for that droplet, to the nearest degree? Create a
table like the one shown and record your results.
Droplet
Original Width Length
A
0.8 cm
B
0.8 cm
droplet E
16
Use the diagram below to determine a relationship to calculate the
angle of impact, u, given a droplet of width AB.
15
A.
14
How far above the ruler, to the nearest centimetre, did each
droplet originate?
13
?
sin u
u
Checklist
Task
✔ Did you draw correct
sketches for the problem?
✔ Did you show your work?
✔ Did you provide appropriate
reasoning?
✔ Did you explain your
thinking clearly?
C.
Measure the length of the other four droplets. Use your relationship
from part A to determine u for each droplet.
D.
If a droplet strikes the ruler head-on, the droplet originated from
distance h. If the droplet strikes the ruler at some angle of impact, the
length of the droplet on the ruler would be d. Use the diagram at the
right to determine the distance h for droplets A to E.
E.
NEL
How far above the ruler did each droplet originate? Explain what
happened with droplet B. Where appropriate, round to the nearest
tenth of a centimetre.
O
h
␪
d
mm
1
2
3
4
5
6
7
8
9 10 11
12 13
317

APPLY the Math

Transcription

Documents pareils

1- Civil engineering is

www.sunex.com DSL203 Digital Imaging Optics Wide Angle

SP2-115 - N.W.T Solutions

Projecteur led blanc chaud 220v 200w angle 140 TLB200WC

Evolution and Original Sin - Washington Theological Consortium

Precise formulas for the distributions of the principal geometric

Finished Size: 54 x 68